Bài 3: Tìm x, biết:

a) \(16x^2-\left(4x-5\right)^2=15\)

\(\Leftrightarrow16x^2-16x^2+40x-25-15=0\)

\(\Leftrightarrow40x-40=0\)

\(\Leftrightarrow4x=40\)

\(\Leftrightarrow x=10\)

Vậy x = 10

b) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(\Leftrightarrow\left(2x+3\right)^2-4\left(x^2-1\right)=49\)

\(\Leftrightarrow4x^2+12x+9-4x^2+4-49=0\)

\(\Leftrightarrow12x-36=0\)

\(\Leftrightarrow12x=36\)

\(\Leftrightarrow x=3\)

Vậy x = 3

c) \(\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2=18\)

\(\Leftrightarrow\left(1-2x\right)\left(2x+1+1-2x\right)=18\)

\(\Leftrightarrow2\left(1-2x\right)=18\)

\(\Leftrightarrow2-4x=18\)

\(\Leftrightarrow4x=-16\)

\(\Leftrightarrow x=-4\)

Vậy x =-4

d) \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\)

\(\Leftrightarrow12x-5=0\)

\(\Leftrightarrow12x=5\)

\(\Leftrightarrow x=\frac{5}{12}\)

Vậy \(x=\frac{5}{12}\)

e) \(\left(x-5\right)^2-x\left(x-4\right)=9\)

\(\Leftrightarrow x^2-10x+25-x^2+4x=9\)

\(\Leftrightarrow25-6x=9\)

\(\Leftrightarrow6x=16\)

\(\Leftrightarrow x=\frac{8}{3}\)

Vậy \(x=\frac{8}{3}\)

f) \(\left(x-5\right)^2+\left(x-4\right)\left(1-x\right)=0\)

\(\Leftrightarrow x^2-10x+25+x-x^2-4+4x=0\)

\(\Leftrightarrow21-5x=0\)

\(\Leftrightarrow5x=21\)

\(\Leftrightarrow x=\frac{21}{5}\)

Vậy \(x=\frac{21}{5}\)

bài của bạn làm sai rồi :)

a) \(\left(2x+1\right)\left(1-2x\right)+\left(2x-1\right)^2=22\)

\(\Rightarrow\left(1+2x\right)\left(1-2x\right)+\left[\left(2x\right)^2-2.2x+1^2\right]=22\)

\(\Rightarrow1^2-\left(2x\right)^2+\left(4x^2-4x+1\right)=22\)

\(\Rightarrow1-4x^2+4x^2-4x+1=22\)

\(\Rightarrow2-4x=22\)

\(\Rightarrow-4x=22-2=20\)

\(\Rightarrow x=20:\left(-4\right)=-5\)

b/ \(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2.\left(x+1\right)^2=0\)

\(\Rightarrow\left(x^2-2.x.5+5^2\right)+\left(x^2-3^2\right)+2.\left(x^2+2.x.1+1^2\right)=0\)

\(\Rightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)

\(\Rightarrow x^2-10x+25+x^2-9-2x^2-4x-2=0\)

\(\Rightarrow-14x+14=0\)

\(\Rightarrow-14x=0-14=-14\)

\(\Rightarrow x=\left(-14\right):\left(-14\right)=1\)

b/\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)

\(\Leftrightarrow x^2-10x+25+x^2-3^2-2\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow x^2-10x+25+x^2-9-2x^2-4x-2=0\)

\(\Leftrightarrow14x=14\Leftrightarrow x=1\)

c/\(\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(4x^2-9\right)=0\)

\(\Leftrightarrow4x^2+12x+9+4x^2-12x+9-8x^2+18=0\)

\(\Leftrightarrow0x=-36\Leftrightarrow x=0\)

a/\(\left(2x+1\right).\left(1-2x\right)+\left(2x-1\right)^2=22\Leftrightarrow2x-4x^2+1-2x+4x^2-4x+1=22\Leftrightarrow-4x=20\Leftrightarrow x=-5\)

\(a)\)

\(f\left(x\right)=2x.\left(x^2-3\right)-4.\left(1-2x\right)+x^2.\left(x-2\right)+\left(5x+3\right)\)\(=2x^3-6x-4+8x+x^3-2x^2+5x+3=3x^3+7x-1-2x^2=3x^3-2x^2+7x-1\)\(g\left(x\right)=-3.\left(1-x^2\right)-2.\left(x^2-2x-1\right)=-3+3x^2-2x^2+4x+2=-1+x^2+4x=x^2+4x-1\)

\(b)\)

\(h\left(x\right)=f\left(x\right)-g\left(x\right)=\left(3x^3-2x^2+7x-1\right)-\left(-1+x^2+4x\right)=x^2+4x-1=3x^3-2x^2+7x-1+1-x^2-4x=3x^3-3x^2+3x\)

\(\text{Xét}:\)

\(3x^3-3x^2+3x=0\)

\(\rightarrow3x.\left(x^2-x+1\right)=0\)

\(\rightarrow x.\left(x^2-x+1\right)=0\)

\(\rightarrow\orbr{\begin{cases}3x.\left(x^2-x+1\right)=0\\x.\left(x^2-x+1\right)=0\end{cases}}\)     \(\rightarrow\orbr{\begin{cases}x=0\\x^2-x+1=0\end{cases}}\)

\(\rightarrow\orbr{\begin{cases}x=0\\x\notinℝ\end{cases}}\)                                   \(\rightarrow x=0\)

\(\text{Vậy nghiệm của}\)\(h\left(x\right)\)\(\text{là}:\)\(0\)

a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+2=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-2}{3}\\x=5\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)

b. \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)

\(\Leftrightarrow3x\left(x-4\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

\(\Rightarrow S=\left\{0;4\right\}\)

c. \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)

\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)

\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)

d. \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)

\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)

e. \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(6x+1-x+2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)

g. \(\left(2x-1\right)^2=49\)

\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)

Bài 1: Tìm x

a) 2x - 15 = -27
b) 2 (x + 1) – 3 = 7

c) 14 – (40 – x) = -27
d) 96 – 2(4 – 5x) = -12

e) – 40 – (– 3 – 33) + (40 – x) = – (– 47)
f) x(3x – 9). (121 – x2) = 0

g) – 62 – (38 + x) + 2x = – 100
h) (x + 1)2.(x2 + 1) = 0

i) (x – 12) – (2x + 31) = 6
k) 17/ (x + 3)3 : 3 – 1 = – 10

Bài 1: a) 2x - 15 = 27 => 2x = 27 + 15 = 42 => x = 42 : 2 = 21. b) 2(x + 1) - 3 = 7 => 2(x + 1) = 7 + 3 = 10 => x + 1 = 10 : 2 = 5 => x = 5 - 1 = 4. c) 14 - (40 - x) = -27 => 40 - x = 14 - (-27) = 41 => x = 40 - 41 = -1. d) 96 - 2(4 - 5x) = -12 => 2(4 - 5x) = 96 - (-12) = 108 => 4 - 5x = 108 : 2 = 54 => 5x = 4 - 54 = -50 => x = (-50) : 5 = -10. 

e) (-40) - [(-3) - 33] + (40 - x) = -(-47) => (-40) - (-36) + 40 - x = 47 => (-40) + 36 + 40 - x = 47 => 36 - x = 47 => x = 36 - 47 = -11. f) x(3x - 9).(121 - 2x) = 0 => x hoặc (3x - 9) hoặc (121 - 2x) bằng 0 => ta có 3 TH: TH1: x = 0 ; TH2: 3x - 9 = 0 => 3x = 0 + 9 = 9 => x = 9 : 3 = 3 ; TH3: 121 - 2x = 0 => 2x = 121 - 0 = 121 (vô lý)(loại). Vậy x ∊ {0;3} 

a: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

nên \(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{6}{10}=\dfrac{3}{5}\)

hay \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b: \(\Leftrightarrow5-\dfrac{4}{7}x=13\)

nên 4/7x=-8

hay x=-12

c: \(\left(x+\dfrac{1}{2}\right)\cdot\left(\dfrac{2}{3}-2x\right)=0\)

=>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

d: \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

nên 1/6x=5/12

hay x=5/2

3 . (2x - 1) + (x - 4) = 0

➤ (2x - 1) + (x - 4) = 0

⇒ \(\left[{}\begin{matrix}2x-1=0\\x-4=0\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=0,5\\x=4\end{matrix}\right.\)

Vậy 3 (2x - 1) + (x - 4) = 0

= 3 (2 . 0,5 - 1) + (4 - 4) = 0

3 . (2x - 1) + (x - 4) = 0

(2x - 1) + (x - 4) = 0

⇒

⇒

Vậy 3 (2x - 1) + (x - 4) = 0

= 3 (2 . 0,5 - 1) + (4 - 4) = 0

a) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)

\(\Leftrightarrow-12x=9+16\Leftrightarrow-12x=24\Leftrightarrow x=-2\)

Vậy x=-2

b) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x=1-32-9\Leftrightarrow2x=-4x\Leftrightarrow x=-20\)

Vậy x=-20

c) \(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\)

\(\Leftrightarrow8x=36-63-1-12\Leftrightarrow8x=-40\Leftrightarrow x=-5\)

Vậy x=-5

d) \(\Leftrightarrow x^3-27+x\left(4-x^2\right)=1\)

\(\Leftrightarrow x^3-27+4x-x^3=1\Leftrightarrow4x=1+27\Leftrightarrow4x=28\Leftrightarrow x=7\)

Vậy x=7

e) \(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-19\)

\(\Leftrightarrow12x=-19+6-1-1\Leftrightarrow12x=-15\Leftrightarrow x=-\frac{5}{4}\)

Vậy x= -5/4